What is the answer to 3^{2x}=4^{(x+1)} ?

The answer to 3^{2x}=4^{(x+1)} is x=(ln(2))/(ln(3)-ln(2))

Solve for x: 3^{2x}=4^{(x+1)}

The solution is x=(ln(2))/(ln(3)-ln(2))

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3^{2 x} = 4^{(x + 1)}

x = \frac{ln ( 2 )}{ln ( 3 ) - ln ( 2 )}

Decimal Notation

x = 1.70951 \dots

Solution steps

3^{2 x} = 4^{(x + 1)}

Apply exponent rules

2 x ln (3) = (x + 1) ln (4)

Solve

2 x ln (3) = (x + 1) ln (4) : x = \frac{ln ( 2 )}{ln ( 3 ) - ln ( 2 )}

x = \frac{ln ( 2 )}{ln ( 3 ) - ln ( 2 )}

15 = 17 e^{- (0.116) t}

2^{x + 1} + 2^{x - 2} = \frac{9}{2}

2^{x + 3} - 2^{x} = 112

x \times 2^{x - 1} = 32

3^{x + 1} + 3^{x} + 3^{x + 1} = 39

What is the answer to 3^{2x}=4^{(x+1)} ?
The answer to 3^{2x}=4^{(x+1)} is x=(ln(2))/(ln(3)-ln(2))
Solve for x: 3^{2x}=4^{(x+1)}
The solution is x=(ln(2))/(ln(3)-ln(2))